Problem: $\int \dfrac{6x}{(5+3x^2)^4}\,dx\,= $ $+~C$
Answer: Notice that we can rewrite the integral as $ \int \dfrac{1}{(5+3x^2)^4}\cdot \,6x\, dx\,$. If we let $ {u=5+3x^2}$, then $du=6x \, dx}$. Substituting gives us: $ \int \dfrac{1}{({5+3x^2})^4}\cdot \,6x\, dx}\,= \int \dfrac{1}{ u^4}\, du}\,$ We recognize this antiderivative. $\begin{aligned}\phantom{\int\dfrac{e^x}{1+e^{2x}}dx}&= \int\dfrac{1}{u^4}\,du~\\\\\\ &=-\dfrac{1}{3u^3}+C\end{aligned}$ We can now substitute back to find the antiderivative in terms of $x$. ∫ e x 1 + e 2 x d x = − 1 3 u 3 + C = − 1 3 ( 5 + 3 x 2 ) 3 + C \begin{aligned}\phantom{\int\dfrac{e^x}{1+e^{2x}}dx~}&=-\dfrac{1}{3u^3}+C\\\\\\\ &=-\dfrac{1}{3(5+3x^2)^3}+C\end{aligned} The answer: $\int \dfrac{6x}{(5+3x^2)^4}\,dx\,= -\dfrac{1}{3(5+3x^2)^3}+C$